Problem: Simplify and expand the following expression: $ \dfrac{t + 7}{5t + 8}-\dfrac{4t - 7}{t + 7} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(5t + 8)(t + 7)$ Multiply the first term by $\dfrac{t + 7}{t + 7}$ $ \begin{align*} \dfrac{t + 7}{5t + 8} \times \dfrac{t + 7}{t + 7} & = \dfrac{(t + 7)(t + 7)}{(5t + 8)(t + 7)} \\ & = \dfrac{t^2 + 14t + 49}{(5t + 8)(t + 7)}\end{align*} $ Multiply the second term by $\dfrac{5t + 8}{5t + 8}$ $ \begin{align*} \dfrac{4t - 7}{t + 7} \times \dfrac{5t + 8}{5t + 8} & = \dfrac{(4t - 7)(5t + 8)}{(t + 7)(5t + 8)} \\ & = \dfrac{20t^2 - 3t - 56}{(t + 7)(5t + 8)}\end{align*} $ Now we have: $ = \dfrac{t^2 + 14t + 49}{(5t + 8)(t + 7)} - \dfrac{20t^2 - 3t - 56}{(t + 7)(5t + 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{t^2 + 14t + 49 - (20t^2 - 3t - 56)}{(5t + 8)(t + 7)} $ $ = \dfrac{t^2 + 14t + 49 - 20t^2 + 3t + 56}{(5t + 8)(t + 7)} $ $ = \dfrac{-19t^2 + 17t + 105}{(5t + 8)(t + 7)}$ Expand the denominator: $ = \dfrac{-19t^2 + 17t + 105}{5t^2 + 43t + 56}$